Extending Rolle's Theorem to Infinity

Consider $$ g(x)=f(\tan x) $$ defined over $(-\pi/2,\pi/2)$. Then we can extend it to $[-\pi/2,\pi/2]$ by setting $$ g(-\pi/2)=g(\pi/2)=L $$ The derivative of $g$ exists over $(-\pi/2,\pi/2)$ and is equal to $$ g'(x)=\frac{f'(\tan x)}{\cos^2x} $$ By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(\tan c)=0$.

You can do similarly for a function $f$ defined over $[a,\infty)$, differentiable over $(a,\infty)$ and such that $$ f(a)=\lim_{x\to\infty}f(x) $$ Define $g(x)=f(\tan x)$ over $[\arctan a,\pi/2)$ and the proof will be the same as before.

Derivatives satisfy the intermediate value property.

If $f'(c) \not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.

In either case, $f$ is strictly monotone ruling out $\displaystyle \lim_{x \to -\infty} f(x) = \lim_{x \to \infty} f(x)$.